Independent adjustment of bass and treble
frequencies in high fidelity audio amplifiers is usually accomplished
utilizing specially designed tone-control networks. There are versions for
these tone controls based only on passive components, such as the
ground-referenced James network shown in fig. 1. Among those versions using
active devices we must make mention of P.J. Baxandall´s proposal, in which the
tone control was devised as a feedback amplifier (refs. 1 and 2).

In this article we shall analyze the James network
(also known as the passive Baxandall tone control), obtaining its design
equations.

Let´s begin studying the bass control section
(fig.2), which has influence over the frequencies below the designed center
frequency of the overall James network.

Here, R4 offers some isolation between
this stage and the treble control section (frequencies above the center
frequency are affected by this control). R5 represents the input
resistance of the amplifier connected to the output of the James network and
should be selected such that it imposes no appreciable load on the network.
Here we assume that C3 and C4 are open circuits at the
bass frequencies.

With full bass boost (R2´s slider at the
upper end), the equivalent circuit for the tone control is as depicted by
fig.3.

From this figure and knowing that “s” is Laplace´s
variable, we obtain:

where:

Then:

After some algebraic work we arrive to:

...(1)

The gain for high bass frequencies is:

The gain at low frequencies is:

Expression (1) has a zero given by:

and a pole:

With the tone control adjusted for maximum bass cut
(R2´s slider at the lower end), the equivalent circuit changes to
that of fig.4. Again, we neglect the loading effect of R4 and R5.

Now the following holds:

A simple algebraic manipulation takes us to:

...(2)

The gain at high bass frequencies is now given by:

For low frequencies the gain is:

Expression (2) has a zero given by:

and a pole:

The gain ratio at low frequencies is:

For a 40dB control range the following relationship
must be satisfied:

...(3)

Then:

On the other hand, at maximum bass boost, the ratio
of the gains for low and high bass frequencies is:

According to expression (3):

Then:

For a 20dB bass boost:

Solving for R1 we arrive to the following
relationship:

We may check that:

this is, the ratio of the high-frequency and
low-frequency bass gains is also 20dB at maximum bass cut.

For best symmetry in the response curves we must
choose s01 = sp4. Therefore:

Substituting the already obtained relationships
between the resistances yields:

Then, C2 = 10C1.

The corresponding Bode plot of the bass response at
full boost and full cut is shown below.

Next, we will analyze the treble control. We have already stated
that the treble frequencies are those above the center frequency for which the
James network is designed. We may consider C1 and C2 as
short circuits at these frequencies. Therefore, R1 and R3
along with R4 are part of the treble network (fig.6).

In order to facilitate calculations, we shall first
find the Thevenin equivalent for Vg, R1 and R3.
In fig.7, the equivalent Thevenin voltage source, V1, is given by:

...(4)

The equivalent Thevenin resistance, RTH,
is:

With the treble control adjusted for maximum boost
(R6´s slider at the upper end), and assuming the current through
resistor R6 is much smaller than that flowing through R4,
we may state the following:

if R5>>R4.

Substituting the value given by expression (4) for V1:

Rearranging this equation we get:

Then:

...(5)

At sufficiently low treble-frequencies:

the same as -20.83dB.

At sufficiently high treble-frequencies:

or 0dB.

Expression (5) has a zero given by:

and a pole:

With the treble control adjusted for maximum cut (R6´s
slider at the lower end) we may write:

Then:

From the above equation and bearing in mind (4):

...(6)

At sufficiently low treble-frequencies:

or -20.83dB.

At sufficiently high treble-frequencies:

Expression (6) has a pole given by:

For best symmetry in the response curves we must
choose sp7 = s05. Therefore:

yielding C4 = 11C3.

Fig.8 shows the corresponding Bode plot of the
treble response at full boost and full cut.

A convenient value for R6 must now be
computed. We need some knowledge on the constraints the network imposes before
we may neglect the current through R6, as compared to that
circulating through R4 (IR6<<IR4). IR4
is given by the following expression when the treble control is adjusted for
full cut:

and IR6 by:

Therefore, the requirement that must be satisfied
is:

The worst case for the above inequality occurs when s
approaches infinity. Translated to our analysis, at sufficiently high
treble-frequencies. Then, it is required that:

or equivalently:

At sufficiently high frequencies:

It is clear then that the following must be
satisfied:

Then:

...(7)

If we conduct a similar analysis when the treble
control is at full boost, we would find that the condition to be met is R6>>1.1R4.
Hence, expression (7) prevails.

We would like to arrive at easy-to-use design
formulae, so, before presenting a design example for the James network we shall
try to simplify the expressions for f01, fp2, f03,
fp4, f05, fp6 and fp7. First, for
convenience, the relationships between the components’ values will be repeated.
These are:

R1 = 10R3

R2 = 99R3

C2 = 10C1

C4 = 11C3

f01 is given by:

...(8)

fp2is given by:

...(9)

f03is given by:

fp4 = f01, required for the
desired symmetry on the response curves.

f05 is given by:

...(10)

fp6is given by:

fp7 = f05, required
for the desired symmetry on the response curves.

The center frequency of the James network is taken
as the geometric mean of f01and f05, this
is:

...(11)

which coincides with the frequency of the minimum of
the amplitude response curve when the bass and treble controls are at full
boost. Agrees also with the frequency of the maximum of the amplitude response
curve when both controls are at full cut. Usually, 1kHz is adopted as the
center frequency.

A Design Example

Let us suppose we wish to design a tone control for
a transistorized piece of equipment. An adequate value for R1 is
10k ohms. R3 will then be a 1k ohm-resistor and R2 a 100k
ohm-potentiometer (standard value).

It is convenient that f01 and f05
be separated one decade in frequency. Then, from expression (11) we may obtain
that:

Being fc = 1kHz, we find that f01
must be 316Hz. The value for f05will then be 3.16kHz.

From expression (8) we obtain for C2 a
value of 503.65nF. Then, C1 should have a capacitance of 50.36nF.
According to (9), fp2 = 31.6Hz.

R5 is taken equal to 5 times R2
in order to avoid loading effects on the bass network. Therefore, R5
= 500k ohms. R4 must be chosen such that an affordable value for R6
is obtained when using inequality (7). If we make R4 = 5k ohms, then
R6 must satisfy the condition R6>>65k ohms. We may
adopt a value of 500k ohms for R6.

From expression (10), with the value chosen for R4
we obtain C3 = 774.85pF and C4 = 8.52nF.

Finally, in order to avoid additional attenuation in
the circuit, the output resistance of generator Vg (source resistance)
must be made some 20 times smaller than R1.

Simulation of the frequency response of the James
network

Tone Stack Calculator 1.3 is an excellent software
program that can be used for simulation of tone-control networks. It may be
downloaded from:

Two simulations have been made using this program.
The first simulation uses the above calculated component values and the second
one, standard capacitor values for the circuit. No major variation on frequency
response has been observed between them. Simulation results are shown below.

Fig.9 Frequency response of the James network with
the bass and treble controls at full boost (upper) and full cut (lower).

Fig.10 Frequency response of the James network when
standard capacitance values for C1, C2, C3 and
C4 are used.